# Is there a universal mapping property satisfied by the disjoint union of totally ordered sets?

It is proved in another post that the product and coproduct do not exist in the category of totally ordered sets (except in some trivial cases). (In this post I will only consider the category TOrd, so that the morphisms are the monotone maps, not the strictly monotone maps.)

It seems to me nevertheless that there is a natural way to define the "coproduct" and "product" of totally ordered sets $A$ and $B$ if you do not insist on it being symmetric. In other words, we will have $A \oplus B \not \cong B \oplus A$ and $A \times B \not \cong B \times A$.

• For an asymmetric direct sum (coproduct), let $A \oplus B$ be the disjoint union of $A$ and $B$, where we define $a \le b$ for all $a \in A$, $b \in B$.

• For an asymmetric direct product, let $A \times B$ be the cartesian product of ordered pairs, and let $(a, b) \le (a', b')$ whenever $a < a'$ OR $a = a'$ and $b \le b'$. (I.e., the order is lexicographic).

Can we come up with a universal mapping property which is satisfied by either of these constructions?

Motivation

I understand that it might seem to be wishful thinking to expect these seeming poorly-behaved constructions to satisfy a category-theoretic property. So here is some motivation for why I think they should satisfy such a property.

1. A similar asymmetric construction works in other categories as well. In the category of well-ordered sets, the exact same construction is used for sum and product in ordinal arithmetic. In the category of monoids, the disjoint union of $A$ and $B$ can be a monoid as well if you say that $ab = b$ for any $a \in A$, $b \in B$. The construction agrees with the coproduct and product in the category of sets, although in this case of course both are commutative.

2. Both of these definitions would naturally generalize to taking a direct sum or product of any finite or infinite number of tosets.

3. We have associativity: $A \oplus (B \oplus C) \cong A \oplus (B \oplus C)$ and similarly for $\times$.

4. We also have $A \oplus 0 \cong 0 \oplus A \cong A$ and $1 \times A \cong A \times 1 \cong A$, where $0$ and $1$ are the initial and terminal objects in TOrd.

Disclaimer

My understanding of category theory is mostly limited to its applications in algebra, in particular in groups, commutative rings, and modules.

• You can't hope for a condition phrased in purely category-theoretic language. Namely, there is a functor $T$ from TOrd to itself that consists of inverting all the orderings, and this is actually an isomorphism -- but it doesn't preserve your $\oplus$. Instead $T(A\oplus B)=T(B)\oplus T(A)$. So there's no way, by speaking solely about objects and morphism, to distinguish between whether the left or the right operand to $\oplus$ gets to be on top. – Henning Makholm Dec 29 '14 at 2:44
• @HenningMakholm Interesting. In that case, I wonder if there is a condition that is satisfied both by $A \oplus B$ and by its natural reverse ordering, say $A \oplus' B$, where $\oplus'$ means the other set is on top. Hmm. – 6005 Dec 29 '14 at 7:13

You can't hope for a condition phrased in purely category-theoretic language. Namely, there is a functor $T$ from TOrd to itself that consists of inverting all the orderings, and this is actually an isomorphism -- but it doesn't preserve your $\oplus$. Instead $T(A\oplus B)=T(B)\oplus T(A)$. So there's no way, by speaking solely about objects and morphisms, to distinguish between whether the left or the right operand to $\oplus$ gets to be on top.

However, if we relax the expectation of being purely category-theoretic, we can sort of wing it:

Let $\mathbf 1$ be a terminal object in our category. (Concretely in TOrd, $\mathbf 1$ is an order with one element). For each object $A$, let $0_A$ be the unique morphism $A\to 1$.

Let $\mathbf 2$ be an object such that there are exactly two morphisms $\mathbf 1 \to\mathbf 2$. (Concretely, a two-element order). Choose fixed names for these two morphisms -- for example, call them $\uparrow$ and $\downarrow$. (This is the step that isn't necessarily preserved by isomorphisms of the category).

Now $A\oplus B$ will be an object together with morphisms $f_1:A\to A\oplus B$, $f_2: B\to A\oplus B$, $g: A\oplus B\to\mathbf 2$, such that $g\circ f_1 = {\downarrow}\circ 0_A$ and $g\circ f_2 = {\uparrow}\circ0_B$.

This in itself doesn't specify $A\oplus B$ fully -- concretely, the risk is, first, that it might contain element's hit by neither $f_1$ nor $f_2$, second, that $f_1$ or $f_2$ might not be embeddings.

We could take care of the latter risk by requiring that $f_1$ and $f_2$ be mono. For the former we can require that $f_1$ and $f_2$ are jointly epi: For any $h_1,h_2:A\oplus B\to C$, if $h_1\circ f_1=h_2\circ f_1$ and $h_1\circ f_2=h_2\circ f_2$, then $h_1=h_2$.

However, these local conditions on $f_1$ and $f_2$ don't guarantee (in an arbitrary category with $\mathbf 1$ and $\mathbf 2$) that $A\oplus B$ is unique up to isomorphism, so it is better to use a proper universal property instead:

Whenever we have morphisms $f'_1: A\to C, f'_2: B\to C, g': C\to\mathbf 2$ with $g'\circ f'_1 = {\downarrow}\circ 0_A$ and $g'\circ f'_2 = {\uparrow}\circ 0_B$, there must be a unique $h:A\oplus B\to C$ such that $f'_1 = h\circ f_1$ and $f'_2= h\circ f_2$.

It is tempting also to require $g'\circ h=g$, but it doesn't seem like that is necesary in TOrd (or any concrete category where $\bf 2$ is a two-element set), and it would make the entire construction look less like a colimit. It might be better to view $g$ and $g'$ as "local" features of $A\oplus B$ and $C$, and ignore the fact that both have the same codomain.

A possible problem is that $\bf 2$ is not necessarily unique (though it is in TOrd). But since we're fixing arbitrary choices of the morphisms into $\bf 2$, having to choose $\bf 2$ itself first is arguably not much worse. For example, in the category of partial orders, we can choose $\bf 2$ either as relating its two elements or not; according to that choice $A\oplus B$ will either stack its two arguments atop each other or put them side-by-side as unrelated.

• Isn't $1_A$ the usual symbol for the identity map from $A$ to itself? – Asaf Karagila Dec 29 '14 at 15:17
• @AsafKaragila: Hmm, yes, bad notation. Let me see if I can find something better. – Henning Makholm Dec 29 '14 at 15:20
• +1, thanks for this answer. I've been thinking about it and unfortunately, it seems like my "direct product" is much harder (the obvious projection maps aren't even morphisms!). But this property at least works and may generalize to other categories. – 6005 Jan 4 '15 at 2:55
• @Goos: Worse yet, ${-}\times B$ is not even a functor. – Henning Makholm Jan 5 '15 at 11:18